IB Physics Option
Relativity – Worked Solutions
th
Problems - Relativistic Addition of Velocities (from Giancoli 6 edition)
43. (I) A person on a rocket traveling at 0.50
0.50c (with respect to the Earth) observes a meteor come from
behind and pass her at a speed she measures as 0.50
0.50c.. How fast is the meteor moving with respect to
the Earth?
0.50c with respect to Earth.
44. (II) Two spaceships leave Earth in opposite directions, each with a speed of 0.50
(a)) What is the velocity of spaceship 1 relative to spaceship 2? ((b)) What is the velocity of spaceship 2
relative to spaceship 1?
0.71c.. A second spaceship leaves the first at a speed of 0.87c
45. (II) A spaceship leaves Earth traveling at 0.71
with respect to the first. Calculate the speed of the second ship with respect to Earth if it is fired (a)
( in
the same direction the first spaceship is already moving, ((b)) directly backward toward Earth.
46. (II) An observer on Earth sees an alien ve
vessel approach at a speed of 0.60c. The Enterprise comes to the
rescue (Fig. 26–12),
12), overtaking the aliens while moving directly toward Earth at a speed of 0.90c
0.90 relative
to Earth. What is the relative speed of one vessel as seen by the other?
47. (II) A spaceship in distress sends out two escape pods in opposite directions. One travels at a speed
v1 = −0.60c in one direction, and the other travels at a speed v 2 = +0.70c in the other direction, as
observed from the spaceship. W
What
hat speed does the first escape pod measure for the second escape
pod?
0.75c.. At the same time, rocket B passes Earth moving 0.95c
0.95
48. (II) Rocket A passes Earth at a speed of 0.75
relative to Earth in the same direction. How fast is B moving relative to A when it passes A?
Answers
43. We find the speed in the frame of the Earth from
u=
( v + u′)
vu ′ 

1 + 2 
c 

=
( 0.50c + 0.50c )
=
1 + ( 0.50 ))( 0.50 ) 
0.80c.
44. (a) In the reference frame of the second spaceship, the Earth is moving at 0.50
0.50c,, and the first
spaceship is moving at 0.50cc relative to the Earth. Thus the speed of the first spaceship
relative to the second is
u=
( v + u′)
vu ′ 

1 + 2 
c 

=
( 0.50c + 0.50c )
=
1 + ( 0.50 )( 0.50 ) 
0.80c.
(b) In the reference frame of the first spaceship, the Earth is moving at − 0.50c, and the second
spaceship is moving at − 0.50
0.50c relative to the Earth. Thus the speed of the second spaceship
relative to the first is
u=
−
− 0.50c + ( − 0.50c ) 
= 
= − 0.80c, as expected.
vu ′  1 + ( − 0.50 )( − 0.50 ) 

1
+


c2 

( v + u′)
IB Physics Option
Relativity – Worked Solutions
45. We take the positive direction in the direction of the first spaceship.
(a) In the reference frame of the Earth, the first spaceship is moving at + 0.71c, and the second
spaceship is moving at + 0.87c relative to the first. Thus the speed of the second spaceship
relative to the Earth is
u=
( v + u′ )
vu ′ 

1 + 2 
c 

=
( + 0.71c + 0.87c )
=
1 + ( 0.71)( 0.87 ) 
0.98c.
(b) In the reference frame of the Earth, the first spaceship is moving at + 0.71c, and the second
spaceship is moving at − 0.87 c relative to the first. Thus the speed of the second spaceship
relative to the Earth is
u=
+
+ 0.71c + ( − 0.87 c ) 
=
= − 0.42c.
vu ′  1 + ( 0.71)( − 0.87 ) 

1
+


c2 

( v + u′)
46. We take the positive direction in the direction of the Enterprise.. In the reference frame of the alien vessel,
the Earth is moving at − 0.60c, and the Enterprise is moving at + 0.90c relative to the Earth. Thus the
speed of the Enterprise relative to the alien vessel is
u=
( v + u′)
vu ′ 

1 + 2 
c 

=
( − 0.60c + 0.90c )
=
1 + ( − 0.60 )( + 0.90 )
0.65c.
Note that the relative speed of the two vessels as seen on Earth is 0.90c − 0.60c = 0.30c.
47. From the first pod’s point of view, the spaceship and the second pod are moving in the same direction. Thus
the speed of the second pod relative to the first is
u=
v + u′
0.60c + 0.70c
=
= 0.92c.
vu ′ 1 + ( 0.60 )( 0.70 )
1+ 2
c
48. The speed of rocket B relative to Earth can be written as
u=
v + u′
,
vu ′
1+ 2
c
where v is the speed of Rocket A relative to Earth and u′ is the speed of Rocket B relative to Rocket A.
With the given values, then,
0.95c =
0.75c + u ′
, which gives u ′ = 0.70c.
0.75u ′
1+
c
IB Physics Option
Relativity – Worked Solutions
Special Relativity Part 2: Conceptual Questions from Giancoli
8.
A young-looking
looking woman astronaut has just arrived home from a long trip. She rushes up to an old
gray-haired
haired man and in the ensuing conversation refers to him as her son. How might this be possible?
If the young-looking
looking astronaut has been traveling at extremely high speeds (relative to Earth) for many
Earth years, since her clock will run slow com
compared
pared to the clocks at rest on Earth, it is possible that her son
could now be much older than she is after her trip. This is a twist on the twin paradox.
9.
If you were traveling away from Earth at speed 0.5
0.5c,, would you notice a change in your heartbeat?
Would your mass, height, or waistline change? What would observers on Earth using telescopes say
about you? You would not notice a change in your own heartbeat, mass, height or waistline. No matter
how fast you are moving relative to Earth, you are at re
rest
st in your own reference frame. Thus, you would
not notice any changes in your own characteristics. To observers on Earth, you are moving away at 0.5c,
0.5
which gives γ = 1/0.87.
0.87. To these observers, it would appear that your heartbeat has slowed by a factor of
0.87, that your mass has increased by a factor of 1/0.87 = 1.15, and that your waistline has decreased by a
factor of 0.87 (all due to the relativity equations for time dilation, mass increase, and length contraction),
but that your height would be unchanged (since there is no relative motion between you and Earth in that
direction).
10. Do time dilation and length contraction occur at ordinary speeds, say 90 km h ? Time dilation and
length contraction do occur at 90 km/h, but the eff
effects
ects are so small that they are not measurable.
16.
Can a particle of nonzero rest mass attain the speed of light? No, it is not possible for a nonnon
zero rest mass particle to attain a speed of c.. This is because its relativistic mass would increase to ∞ as
v → c , which means it would take an infinite amount of energy to increase the speed of this infinite mass
particle to c.
17. Does the equation E = mc 2 conflict with the conservation of energy principle? Explain. No, E = mc2 does
not conflict with the conservation of energy, it actually completes it. Since this equation shows us that
mass and energy are interconvertible, it says it is now necessary to iinclude
nclude mass as a form of energy in the
analysis of physical processes.
18. If mass is a form of energy, does this mean that a spring has more mass when compressed than when
relaxed? Yes. One way to describe the energy stored in the compressed spring is to say it is a mass
increase (although it would be so small that it could not be measured). This mass will convert back to
energy when the spring is uncompressed.
19. It is not correct to say that “matter can neither be created nor destroyed.” What must we say instead?
Matter and energy are interconvertible (matter can be converted into energy and energy can be converted
into matter), thus we should say “energy can neither be created nor destroyed.”
20. Is our intuitive notion that velocities simply add, com
completely wrong? No, our intuitive notion that
velocities simply add is not completely wrong. Our intuition is based on our everyday experiences and at
these everyday speeds our intuition is correct regarding how velocities add. Our intuition does break
down,
wn, though, at very high speeds, where we have to take into account relativistic effects. Relativity does
not contradict classical mechanics, but it is a more general theory where classical mechanics is a limiting
case.
IB Physics Option
Relativity – Worked Solutions
Problems
20. (I) A certain chemical reaction requires 4.82 × 10 4 J of energy input for it to go. What is the increase in
mass of the products over the reactants?
20. We find the increase in mass from
4.82 × 104 J )
(
∆E
∆m = 2 =
= 5.36 × 10−13 kg.
2
8
c
( 3.00 × 10 m s )
Note that this is so small, most chemical reactions are considered to have mass conserved.
21. (I) When a uranium nucleus at rest breaks apart in the process known as fission in a nuclear reactor, the
resulting fragments have a total kinetic energy of about 200 MeV. How much mass was lost in the
process?
21. We find the loss in mass from
−13
∆E ( 200 MeV ) (1.60 × 10 J MeV )
∆m = 2 =
= 3.56 × 10−28 kg.
2
8
c
( 3.00 × 10 m s )
22. (I) Calculate the rest energy of an electron in joules and in MeV (1 MeV = 1.60 × 10 −13 J).
22. The rest energy of the electron is
2
E = m0 c 2 = ( 9.11 × 10−31 kg )( 3.00 × 108 m s ) = 8.20 × 10−14 J
=
(8.20 × 10
(1.60 × 10
−13
−14
J)
J MeV )
= 0.511MeV.
23. (I) Calculate the rest mass of a proton in MeV c 2 .
23. The rest mass of the proton is
2
−27
8
E (1.67 × 10 kg )( 2.9979 × 10 m s )
938MeV
m0 = 2 =
=
.
−13
2
c
c2
(1.60 × 10 J MeV ) c
24. (I) The total annual energy consumption in the United States is about 8 × 1019 J. How much mass would
have to be converted to energy to fuel this need?
24. We find the necessary mass conversion from
8 × 1019 J )
(
∆E
∆m = 2 =
= 9 × 102 kg.
2
8
c
( 3.00 × 10 m s )
25. (II) How much energy can be obtained from conversion of 1.0 gram of mass? How much mass could
this energy raise to a height of 0.25 km above the Earth’s surface?
25. We find the energy equivalent of the mass from
2
E = mc 2 = (1.0 × 10−3 kg )( 3.00 × 108 m s ) = 9.0 × 1013 J.
If this energy increases the gravitational energy, we have
E = mgh;
IB Physics Option
Relativity – Worked Solutions
9.0 × 1013 J = m ( 9.80
9.80m
m s 2 )( 0.25 × 103 m ) , which gives m = 3.7 × 1010 kg.
26. (II) What is the speed of a particle when its kinetic energy equals its rest energy?
26. If the kinetic energy is equal to the rest energy, we have
2
2
KE = ( m − m0 ) c = m0 c , or m = 2m0 .
We find the speed from
m0
m=
1
2
;
  v 
1 −  2  
  c 
m0
2m0 =
, which gives v = 0.866c.
1
  v2  2
1 −  2  
  c 
2
27. (II) At what speed will an object’s kinetic energy be 25% of its rest energy?
27. If the kinetic energy is 25% of the rest energy, we have
2
2
KE = ( m − m0 ) c = 0.25m0 c , or m = 1.25m0 .
We find the speed from
m=
1.25m0 =
m0
v2
1− 2
c
m0
v2
1− 2
c
;
, which gives v = 0.60c.
28. (II) How much work is required to accelerate a proton from rest up to a speed of 0.997c?
0.997
28. (a) We find the work required from









1

2
2 
W = ∆KE = ( m − m0 ) c = m0 c 
− 1
1 


2
   v  2 

  1 −  c   

      




 


1
= ( 938MeV
938 MeV )  
− 1 = 11.2 × 103 MeV = 11.2GeV
11.2 GeV ( 2.23 × 10−9 J ) .
1 


2 2 
  1 − ( 0.997 )  



